@[169132:naga25french] -- pls. help with the solutions of the above problems.
Some pls solve the above questions..
Nyone pls tell me how to solve this Permutation and Combination problem:
There are five balls of different colours and 5 boxes of the same colours as those of the balls.No. of ways in which the balls one in each box can be placed such that the ball does nt goes to the box of its own colour is
how many 4 digit odd nos. can be formed such that every 3 in the number is followed by a 6?
@ayushbhalotia said: hi !!Pls. help with the detailed sol for the below mentioned probs of POM COM:-
Q1) the crew of an 8 member rowing team is to be chosen from 12 men, of which 3 must row on 1 side and 2 must row on the other side only. find the number of ways of arranging the crew with 4 members on each side ??
Options :- A) 40320 B) 30240 C) 60480 D) 10080 E) None of these
Q2) in how many ways 5 MBAs and 6 CA students can be arranged together so that no 2 MBA students are side by side:-
Options:-a) 7!*6!/2! b) 6!*6! c) 6!*5! d) 11P5 e) 11C5
Q3) there are 8 orators A,B,C,D,E,F,G and H. In how many ways can the arrangements be made so that A comes before B and B comes before C alwaz??
Options: -A) 8!/3! B) 8!/6! C) 5!*3! D) 6!*3! E) 8!/(5!*3!) F) None of these
Q4) if we have to make 7 boys and 7 girls sit around a round table which is numbered, then the number of ways in which it can be done are??
Options - A) 2*7!*7! B) 7!*6! C) 7!*7! D) None of these
Q5) in a building there are 12 floors excluding the ground floor. 9 ppl get in the lift at the ground floor. the lift does not stop at the 1st floor. 2,3,4 ppl alight from the lift on its upward journey, then how many ways then can do so?
Options- A) 11C3* 3! B) 11P3*9C4*5C3 C) 10P3*9C4*5C3 D) 12C3
Q6) 7 different objects are divided among 3 ppl. in how many ways it can be done if one or two of them can get no objects??
Options - A) 15120 B) 2187 C) 3003 D) 792
i am not sure about the 1st one
Q2)first arrange the 6 law students..there will be then 5 spaces between them and two more at extreme right and left..so the 5 mba students have 2 be placed in these 7 places in 7P5 ways and corresponding to each of these arrangemnts the law students will arrange in 6! ways.ans 7P5*6! ie option a.
Q3)They can arrange among themselves in 8! ways out of which in half of them a will come before b.so in 8!/2 arrangements a comes before b.again in these 8!/2 arrangements b comes befre c in 1/3rd of these//so the required permutation is 8!/3!.
Q4)if the boys and girls are allowed to sit any way they wish then the no of arrangements should be 14!i think u hav got the question wrong.however i came across this type of question somewhere where the boys and girls have to sit alternately.if thats the case then the arrangements would be 2*7!*7!(because the boys can sit either in the even numbered or odd numbered chairs alternately with the girls)
Q5)there are 11 floors to get of out of which we have to consider 3 floors where they get off.this can be done in 11p3 ways as the order of the floors in which they get off has to be considered.so the answer is 11p3*9c4*5c3
Q6)the first of the objects can be given in 3 ways,the 2nd object also in 3 ways...the seventh object can also be given to any one of three people..so ans is 3^7=2187 i think.
Q3)They can arrange among themselves in 8! ways out of which in half of them a will come before b.so in 8!/2 arrangements a comes before b.again in these 8!/2 arrangements b comes befre c in 1/3rd of these//so the required permutation is 8!/3!.
Q4)if the boys and girls are allowed to sit any way they wish then the no of arrangements should be 14!i think u hav got the question wrong.however i came across this type of question somewhere where the boys and girls have to sit alternately.if thats the case then the arrangements would be 2*7!*7!(because the boys can sit either in the even numbered or odd numbered chairs alternately with the girls)
Q5)there are 11 floors to get of out of which we have to consider 3 floors where they get off.this can be done in 11p3 ways as the order of the floors in which they get off has to be considered.so the answer is 11p3*9c4*5c3
Q6)the first of the objects can be given in 3 ways,the 2nd object also in 3 ways...the seventh object can also be given to any one of three people..so ans is 3^7=2187 i think.
@rajatt1557 said: how many 4 digit odd nos. can be formed such that every 3 in the number is followed by a 6?
the no can be 36XX or X36X or XXXX form.the last place can be filled in only by odd nos that is 5 ways.in the first form the the third place can be filled in 9 ways(any no except 3).so total of 45 such nos.in 2nd form the first place can be filled in 8 ways.so total 40 such numbers.the last form can be done in 8*9*9*5=3240.so total of 3325 such nos?
@nfsfreak said:
Q4)if the boys and girls are allowed to sit any way they wish then the no of arrangements should be 14!i think u hav got the question wrong.however i came across this type of question somewhere where the boys and girls have to sit alternately.if thats the case then the arrangements would be 2*7!*7!(because the boys can sit either in the even numbered or odd numbered chairs alternately with the girls)
Hi can you please tell me in this scenario why it should't be 13!
@ayushbhalotia said: hi !!Pls. help with the detailed sol for the below mentioned probs of POM COM:-
Q1) the crew of an 8 member rowing team is to be chosen from 12 men, of which 3 must row on 1 side and 2 must row on the other side only. find the number of ways of arranging the crew with 4 members on each side ??
Options :- A) 40320 B) 30240 C) 60480 D) 10080 E) None of these
As 3 of them need to be in a side and 1 in other so reaming are 12-(3+2) = 7
now we want to select 3 from those 7 people so 7C3
among these 3 people 1 will be on a side =3C1
now arrangement for one side is 4! and for other side is 4!
altogether it should be 7C3*3C1*4!*4! = 60480
Please correct me if I am wrong
@[607979:AvishekAdhvaryu]
it will not be 13! because in this the seats are numbered so the cyclic order wont be maintained as each person can sit in any seat number irrespective of the cyclic arrangement.
@ nfsfreak and --- Thanks for the solutions...ya for the 4th question girls and boys sit alternately...@AvishekAdhvaryu --- since the seats are number we wont be able to fix the positions so we will need to tk 14!..
I had posted these ques on 12th but since no1 replied, after a lot of brain storming i had deduced the answers myself..but still thanks a lot guys... :-)
@[607369:nfsfreak] --- if in the belopw mentioned ques if instead of different objects, identical objects were to be distributed then how it wud hv been done??
7 different objects are divided among 3 ppl. in how many ways it can be done if one or two of them can get no objects??
Pls explain the soln for the below mentioned 4 diff cases: -
Case 1:- 7 different objects are divided among 3 ppl. in how many ways it can be done if one or two of them can get no objects??
Case 2:- 7 identical objects are divided among 3 ppl. in how many ways it can be done if one or two of them can get no objects??
Case 3:- 7 different objects are divided among 3 ppl. in how many ways it can be done if any of them get any number of objects??
Case 4:- 7 identical objects are divided among 3 ppl. in how many ways it can be done if any of them get any number of objects??
Prahass attemps a question paper that has 3 sections with 6 questions each. He has to attempt 8 questions with atleast 2 from each section. in how many ways can he attempt the paper?
P.S.-- Shouldn't the answer be 6C2* 6C2*6C2*12C2 ?? Well , it is not . Can anyone explain why? what cases are left in this approach
@ayushbhalotia said:Pls explain the soln for the below mentioned 4 diff cases: -
Case 1:- 7 different objects are divided among 3 ppl. in how many ways it can be done if one or two of them can get no objects??
Case 2:- 7 identical objects are divided among 3 ppl. in how many ways it can be done if one or two of them can get no objects??
Case 3:- 7 different objects are divided among 3 ppl. in how many ways it can be done if any of them get any number of objects??
Case 4:- 7 identical objects are divided among 3 ppl. in how many ways it can be done if any of them get any number of objects??
Case1
7 different objects are to be divided so for 1st object i have 3 choices for 2nd i have 3 choices so total 3*3*3*....7time so 3^7
Case2
7 identical object are to be divided so
A+B+C=7
as you have stated 1 or 2 can get zero so you have to calculate number of non-negative solution
so (7+3-1)C(3-1)=9C2
Case3 is same as case1 as one or two can get no item means anyone can get all item so anybody can get any no of item
If you tell that everyone will get minimum 1 item then 1 goes to each of them 4 remains so each of them can go with 3 ways so total 3^4 ways
Case4 is same as case 2
If you tell that everyone will get minimum 1 item then
A+B+C=7 now as 0 is not possible so we have to calculate no of non-zero solution
so (7-1)C(3-1)=6C2
Please tell me if I am wrong.
first
[ furst] adjective
1.
being before all others with respect to time, order, rank, importance, etc., used as the ordinal number of one: the first edition; the first vice president.
synonyms
ahead, antecedent, basic, beginning, cardinal, early, elementary, front, fundamental, head
@noobster said: Prahass attemps a question paper that has 3 sections with 6 questions each. He has to attempt 8 questions with atleast 2 from each section. in how many ways can he attempt the paper?P.S.-- Shouldn't the answer be 6C2* 6C2*6C2*12C2 ?? Well , it is not . Can anyone explain why? what cases are left in this approach
As the paper contains 3 sections 6 each so no of possible combinations are
(2,2,4),(2,4,2),(4,2,2),(3,3,2),(3,2,3),(3,3,2)
For first 3 total is 3*6C2*6C2*6C4
For last 3 total is 3*6C2*6C3*6C3
total 3*6C2*6C2*6C4+3*6C2*6C3*6C3
@AsihekAdhvaryu said:As the paper contains 3 sections 6 each so no of possible combinations are(2,2,4),(2,4,2),(4,2,2),(3,3,2),(3,2,3),(3,3,2)For first 3 total is 3*6C2*6C2*6C4For last 3 total is 3*6C2*6C3*6C3total 3*6C2*6C2*6C4+3*6C2*6C3*6C3
This approach is correct. I had a doubt in the previously mentioned aproach:
If we select 2 questions from section 1 in 6C2 ways,
then 2 from section 2 and 2 from section 3 in 6c2 ways each, we will be left with 12 unanswered questions with 2 more to do . so we can do it in 12c2 ways.
therefore total ways = 6C2* 6C2*6C2*12C2
But the ans doesn't match with this approach. can u pls explain the discrepancy or error in this approach
If we select 2 questions from section 1 in 6C2 ways,
then 2 from section 2 and 2 from section 3 in 6c2 ways each, we will be left with 12 unanswered questions with 2 more to do . so we can do it in 12c2 ways.
therefore total ways = 6C2* 6C2*6C2*12C2
But the ans doesn't match with this approach. can u pls explain the discrepancy or error in this approach
@[578805:AsihekAdhvaryu]
for Case 3:- 7 different objects are divided among 3 ppl. in how many ways it can be done if any of them get at least 1 object??
According to me the sol shud be that any 3 object are selected and then distributed to these 3 ppl and the remaining 4 objects are then distributed... 7C3 * 3! * (3^4)....
pls. correct me if i am wrong..
@[361226:ayushbhalotia]
Sry for the mistake
You are right
@noobster said:This approach is correct. I had a doubt in the previously mentioned aproach: If we select 2 questions from section 1 in 6C2 ways, then 2 from section 2 and 2 from section 3 in 6c2 ways each, we will be left with 12 unanswered questions with 2 more to do . so we can do it in 12c2 ways. therefore total ways = 6C2* 6C2*6C2*12C2 But the ans doesn't match with this approach. can u pls explain the discrepancy or error in this approach
since 6c2*6c2*6c2 implies selecting 2 questions from each of the three groups and if u multiply 12c2 on top of that it implies that there is a further 4th group with 12 questions out of which we are selecting 2.actually we have to select the 2 remaining questions out of 4 questions either 2 at a time from one particular group or 2 questions one each from two separate groups each of which have 4 questions remaining.
How we do these type of problem?
selecting 25 objects out of 15 of 1 kind,20 of 2nd kind and 25 of 3rd kind???